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4. Systems of Equations & Inequalities
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Algebra 1

4.03 Elimination method

Lesson
Practice

The elimination method

The elimination method is an algebraic method for solving systems of equations where like terms are aligned.

In this system, the equations are aligned:\begin{aligned} 3x+4y&=12\\\ x+y&=11\end{aligned} and in this system, the equations are not aligned: \begin{aligned} 3x&=12-4y\\ x&+y=11\end{aligned}

Elimination method

A method of solving a system of equations by adding or subtracting the equations until only one variable remains

Exploration

Begin by graphing the system of equations and identifying the solution.\begin{aligned} 3x+2y&=10\\x+2y&=8\end{aligned}

Next, perform each operation using the original equations and consider the result. What do you notice? Record your observations.

  1. Divide the second equation by 2 and regraph the system. What do you notice?
  2. Multiply the first equation by 2 and regraph the system. What do you notice?
  3. Add the equations together and regraph the system. What do you notice?
  4. Multiply the first equation by -1 and add the equations together. Graph the new equation with the system. What do you notice?
  5. Multiply the second equation by 3 and add the equations together. Graph the new equation with the system. What do you notice?
  6. Multiply the second equation by -3 and add the equations together. Graph the new equation with the system. What do you notice?

What happens to the solution to a system of linear equations if one equation is multiplied by a number and then added to the other equation?

The goal of the elimination method is to combine the two equations in the system of equations until one variable is eliminated from the problem. When we replace an equation in a system of equations with the sum of that and the multiple of the other, we get a new system with the same solution. This becomes the basis of the elimination method.

We can also determine the number of solution from a system of equations without solving the system. If one equation in the system is a multiple of the other, the system has infinite solutions:

2y - 3 = 4

6y - 9 = 12

This system has infinite solutions because they are different representations of the same line.

If the coefficients of the x and y terms are multiples but the constant is not a multiple, the system has no solutions:

2y - 3 = 4

6y - 9 = 5

This system has no solutions because they represent parallel lines and will never cross.

Lastly, if the x and y terms are not multiples then the lines intersect and have one solution:

2y - 3 = 4

6y - 6 = 5

This system represents lines that intersect exactly once, so there will be a single solution.

Examples

Example 1

Consider the system of equations: \begin{aligned} 4x+y&=11 \\x-2y&=5\end{aligned}

a

Equation 1 is multiplied by 2 to produce the system: \begin{aligned} 8x+2y&=22 \\x-2y&=5\end{aligned} Determine whether or not the two systems are equivalent. Then decide whether or not the two systems result in the same solution.

Worked Solution
Create a strategy

Two systems are equivalent if they have exactly the same graphs. The two systems have the same solution if they intersecct at the exact same point.

Apply the idea

The systems are equivalent since they share the same graphs. Since the systems are equivalent, they must also intersect at the same point.

-1
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-1
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Reflect and check

These two systems are equivalent because the equations 4x+y=11 and 8x+2y=22 are on the exact same line. Multiplying the entire equation by a constant does not change the set of x and y values that make the equation true.

b

Equation 2 is replaced with the sum of Equation 2 and 3 times Equation 1 to produce the system: \begin{aligned} 4x+y &=11\\13x+y&=38 \end{aligned} Determine whether or not the two systems are equivalent. Then decide whether or not the two systems result in the same solution.

Worked Solution
Create a strategy

Two systems are equivalent if they have the exact same graphs. The systems have the same solution if they intersect at the same point.

Apply the idea

The systems of equations are not equivalent because the graphs are not exactly the same. However, both systems do have the same solution, because they intersect at the same point:

1
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-4
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Reflect and check

We can also solve both systems using substitution to verify that the solutions are equal:

Begin by numbering the equations, so the solving process is easier to explain:

1\displaystyle 4x+y\displaystyle =\displaystyle 11
2\displaystyle x-2y\displaystyle =\displaystyle 5
\displaystyle 4x+y\displaystyle =\displaystyle 11Equation 1
\displaystyle y\displaystyle =\displaystyle 11-4xSubtract 4x from both sides
\displaystyle x-2(11-4x)\displaystyle =\displaystyle 5Substitute y=11-4x into Equation 2
\displaystyle x-22+8x\displaystyle =\displaystyle 5Distributive property
\displaystyle 9x-22\displaystyle =\displaystyle 5Combine like terms
\displaystyle 9x\displaystyle =\displaystyle 27Add 22 to both sides
\displaystyle x\displaystyle =\displaystyle 3Divide both sides by 9
\displaystyle 4\left(3\right)+y\displaystyle =\displaystyle 11Substitute x=3 into Equation 1
\displaystyle 12+y\displaystyle =\displaystyle 11Evaluate the multiplication
\displaystyle y\displaystyle =\displaystyle -1Subtract 12 from both sides

The solution to this system is (3,-1).

For the second system of equations:

1\displaystyle 4x+y\displaystyle =\displaystyle 11
2\displaystyle 13x+y\displaystyle =\displaystyle 38
\displaystyle 4x+y\displaystyle =\displaystyle 11Equation 1
\displaystyle y\displaystyle =\displaystyle 11-4xsubtract 4x from both sides
\displaystyle 13x+(11-4x)\displaystyle =\displaystyle 38Substitute y=11-4x into Equation 2
\displaystyle 9x+11\displaystyle =\displaystyle 38Combine like terms
\displaystyle 9x\displaystyle =\displaystyle 27Subtract 11 from both sides
\displaystyle x\displaystyle =\displaystyle 3Divide both sides by 9
\displaystyle 4\left(3\right)+y\displaystyle =\displaystyle 11Substitute x=3 into Equation 1
\displaystyle 12+y\displaystyle =\displaystyle 11Evaluate the multiplication
\displaystyle y\displaystyle =\displaystyle -1Subtract 12 from both sides

The solution to this system is also (3,-1).

Example 2

Solve each system of equations using the elimination method.

a

\begin{aligned} 9x+y &=62 \\ 5x+y&=38 \end{aligned}

Worked Solution
Create a strategy

Numbering the equations will help us describe our solving process:

1\displaystyle 9x+y\displaystyle =\displaystyle 62
2\displaystyle 5x+y\displaystyle =\displaystyle 38

The goal of the elimination method is to combine the two equations until one variable is eliminated.

In this problem, the y term will be eliminated if we multiply equation 2 by -1 and add it to equation 1.

Apply the idea

Start by multiplying equation 2 by -1:

1\displaystyle 9x+y\displaystyle =\displaystyle 62
2\displaystyle -5x-y\displaystyle =\displaystyle -38

Then, add equation 1 and equation 2 together:

\begin{aligned} &&9x &+y = &62& \\ + &&-5x&-y=&-38& \\ \hline \\ &&4x&+0y=&24& \end{aligned}

\displaystyle 4x\displaystyle =\displaystyle 24Eliminate the y term
\displaystyle x\displaystyle =\displaystyle 6Division property of equality

Now that we know what x equals, we can substitute it back into either of the original equations:

\displaystyle 5\left(6\right) + y\displaystyle =\displaystyle 38Substitute back into equation 2
\displaystyle 30 + y\displaystyle =\displaystyle 38Evaluate the multiplication
\displaystyle y\displaystyle =\displaystyle 8Subtraction property of equality

The solution to this system is the ordered pair \left(6,8\right).

Reflect and check

We could also consider the first two steps as "subtracting the equations". The steps would look like this instead:

\begin{aligned} &&9x &+y = &62& \\ - &&(5x&+y=&38&) \\ \hline \\ &&4x&+0y=&24& \end{aligned}

\displaystyle 4x\displaystyle =\displaystyle 24Eliminate the y term
\displaystyle x\displaystyle =\displaystyle 6Division property of equality
b

\begin{aligned} 2x+3y &= 19 \\ 4x-y&=10\end{aligned}

Worked Solution
Create a strategy

Numbering the equations will help us describe our solving process:

1\displaystyle 2x+3y\displaystyle =\displaystyle 19
2\displaystyle 4x-y\displaystyle =\displaystyle 10

There are multiple combinations that can be used to solve this system:

  • Multiply equation 1 by 2 and subtract it from equation 2
  • Multiply equation 1 by -2 and add it to equation 2
  • Multiply equation 2 by 3 and add it to equation 1

We only need to choose one approach.

Apply the idea

Multiply the first equation by -2:

1\displaystyle -4x-6y\displaystyle =\displaystyle -38
2\displaystyle 4x-y\displaystyle =\displaystyle 10

Then, add equation 1 and equation 2 together:

\begin{aligned} &&-4x &-6y = &-38& \\ + &&4x&-y=&10& \\ \hline \\ &&0x&-7y=&-28& \end{aligned}

\displaystyle -7y\displaystyle =\displaystyle -28Eliminate the x term
\displaystyle y\displaystyle =\displaystyle 4Division property of equality

Now that we know what y equals, we can substitute it back into either of the original equations:

\displaystyle 2x+3\left(4\right)\displaystyle =\displaystyle 19Substitute y=4 into equation 1
\displaystyle 2x+12\displaystyle =\displaystyle 19Evaluate the multiplication
\displaystyle 2y\displaystyle =\displaystyle 7Subtraction property of equality
\displaystyle x\displaystyle =\displaystyle 3.5Division property of equality

The solution to this system is the ordered pair \left(3.5,\,4\right).

Reflect and check

Remember, the solution of the system is unchanged when multiples of one equation are added to the other equation. That means we can substitute our solution for y into either of the original two equations of the system or the resulting equation from when we multiplied the first equation by -2.

c

\begin{aligned} 4x+8y &= 48 \\ 3x-6y&=18\end{aligned}

Worked Solution
Create a strategy

Numbering the equations will help us describe our solving process:

1\displaystyle 4x+8y\displaystyle =\displaystyle 48
2\displaystyle 3x-6y\displaystyle =\displaystyle 18

To make the coefficients of y's the same in both equations, we can multiply the first equation by 3 and the second equation by 4. After that, we can add the equations to eliminate y.

Apply the idea

Multiply the first equation by 3 and the second equation by 4:

1\displaystyle 12x+24y\displaystyle =\displaystyle 144
2\displaystyle 12x-24y\displaystyle =\displaystyle 72

Then, add equation 1 and equation 2 together:

\begin{aligned} &&12x &+24y = &144& \\ + &&12x&-24y=&72& \\ \hline \\ &&24x&-0y=&216& \end{aligned}

\displaystyle 24x\displaystyle =\displaystyle 216Eliminate the y term
\displaystyle x\displaystyle =\displaystyle 9Division property of equality

Now that we know what x equals, we can substitute it back into either of the original equations:

\displaystyle 4\left(9\right)+8y\displaystyle =\displaystyle 48Substitute x=9 into equation 1
\displaystyle 36+8y\displaystyle =\displaystyle 48Evaluate the multiplication
\displaystyle 8y\displaystyle =\displaystyle 12Subtraction property of equality
\displaystyle y\displaystyle =\displaystyle 1.5Division property of equality

The solution to this system is the ordered pair \left(9,1.5\right).

Reflect and check

Let's verify our solution by substituting x = 9 and y = 1.5 back into the original equations.

\displaystyle 4(9)+8(1.5)\displaystyle =\displaystyle 48Substitute into the first equation
\displaystyle 36+12\displaystyle =\displaystyle 48Evaluate the multiplication
\displaystyle 48\displaystyle =\displaystyle 48Evaluate the addition

Thus, our solutions x = 9 and y = 1.5 are correct.

Example 3

When comparing test results, Verna noticed that the sum of her Chemistry and English test scores was 128 and that their difference was 16. She scored higher on her Chemistry test.

a

Write a system of equations for this scenario, where x represents Verna's Chemistry test score, and y represents her English test score.

Worked Solution
Create a strategy

Since we know that Verna's Chemistry test score is higher than her English test score, we should express the difference of the two scores as x - y to get a positive result.

Apply the idea
1\displaystyle x + y\displaystyle =\displaystyle 128
2\displaystyle x - y\displaystyle =\displaystyle 16
Reflect and check

There are two other constraints based on typical test scores that we can apply to this problem:0\leq x \leq 100 and 0\leq y\leq 100.

b

Solve the system of equations to find her test scores.

Worked Solution
Create a strategy

Since the equations have the same coefficient of y with opposite sign, we can add the two equations to eliminate y and solve for x first.

Apply the idea

\begin{aligned} &&x &+y = &128& \\ + &&x&-y=&16& \\ \hline \\ &&2x&+0y=&144& \end{aligned}

\displaystyle 2x\displaystyle =\displaystyle 144Eliminate the y term
\displaystyle x\displaystyle =\displaystyle 72Divide both sides by 2
\displaystyle 72 + y\displaystyle =\displaystyle 128Substitute x=72 into equation 1
\displaystyle y\displaystyle =\displaystyle 56Subtract 72 from both sides

So Verna scored 72 on her Chemistry test and 56 on her English test.

Reflect and check

In this case, the two equations also had the same coefficient of x with the same sign. So we could have instead approached this by subtracting the two equations and solving for y first.

c

Does the solution make sense in terms of the context? Explain your answer.

Worked Solution
Apply the idea

Yes. Assuming that the tests were out of 100, then 72 and 56 are both valid test scores to obtain.

Example 4

Determine the number of solutions to each system without solving.

a

\begin{aligned} 4x+y=11 \\ x-2y=5 \end{aligned}

Worked Solution
Create a strategy

Recall that if one equation in the system is a multiple of the other, the system has infinite solutions because they are different representations of the same line.

If the coefficients of the x and y terms are multiples but the constant is not a multiple, the system has no solutions because they represent parallel lines and will never cross.

Lastly, if the x and y terms are not multiples then the lines intersect and have one solution.

Apply the idea

Since neither the entire equation nor the coefficients of the x and y terms are multiples, these lines intersect exactly once.

Therefore, this system has one solution.

b

\begin{aligned} 2x+3y=-1 \\ 4x+6y=-2 \end{aligned}

Worked Solution
Create a strategy

Recall that if one equation in the system is a multiple of the other, the system has infinite solutions because they are different representations of the same line.

Apply the idea

We can manipulate the second equation by dividing each term by 2:

\displaystyle 4x+6y\displaystyle =\displaystyle -2Original equation
\displaystyle \dfrac{4x}{2}+\dfrac{6y}{2}\displaystyle =\displaystyle \dfrac{-2}{2}Divide each term by 2
\displaystyle 2x+3y\displaystyle =\displaystyle -1

This is exaclty the same as the first equation. Therefore, both equations represent the same line, and there are infinite solutions.

c

\begin{aligned} 2x-4y=0 \\ x-2y=5 \end{aligned}

Worked Solution
Apply the idea

In this system, the coefficients of the x and y terms are multiples but the constant is not a multiple.

Therefore, the system has no solutions because they represent parallel lines and will never cross.

Reflect and check

We can verify this by solving using elimination:

Multiply the second equation by -2:

1\displaystyle 2x-4y\displaystyle =\displaystyle 0
2\displaystyle x-2y\displaystyle =\displaystyle 5

Then, add equation 1 and equation 2 together:

\begin{aligned} &&2x &-4y = &0& \\ + &&-2x&+4y=&-10& \\ \hline \\ &&0x&-0y=&-10& \end{aligned}

\displaystyle 0\displaystyle =\displaystyle -10Eliminate the x and y term

Since the equation 0 = -10 is never true, this system has no solution.

Idea summary

The goal of the elimination method is to combine the equations in a system until one variable is eliminated. We can eliminate a variable in a system of equations by multiplying one (or both) equations so that the coefficients of one variable are equal and opposite.

Outcomes

A.EI.2

The student will represent, solve, explain, and interpret the solution to a system of two linear equations, a linear inequality in two variables, or a system of two linear inequalities in two variables.

A.EI.2a

Create a system of two linear equations in two variables to represent a contextual situation.

A.EI.2b

Apply the properties of real numbers and/or properties of equality to solve a system of two linear equations in two variables, algebraically and graphically.

A.EI.2c

Determine whether a system of two linear equations has one solution, no solution, or an infinite number of solutions.

A.EI.2h

Verify possible solution(s) to a system of two linear equations, a linear inequality in two variable, or a system of two linear inequalities algebraically, graphically, and with technology to justify the reasonableness of the answer(s). Explain the solution method and interpret solutions for problems given in context.

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